e^(i*pi)

[sleeplessdragn]

In my math class a discussion arose concerning this interesting problem. The solution of e^(i*pi) is equal to -1, yet not even my math teacher could explain exactly why this was. All that he could say was that in order to prove this, a derivative of e and the fractional value of pi was involved. I know that it involves a certain trig identity but not much more than that.

After searching through google I found not much more information, although I admit that i did not really google too deeply. I found, though, that a correlation between e and pi has to exist.

Does anyone know of any correlation between e and pi? Are there any practical uses for this equation? Discuss.

[GuidoHunter]

It's been said that that equation is the most magical equation in all of math because it incorporates several different types of math and relates them all. That is, the transcendental, the algebraic, and the imaginary.

I don't know much about it myself, and if you already googled it, I can't really furnish much more.

--Guido

http://andy.mikee385.com

[Kefit]

This equation is incredibly important to physics and differential equations. Lemme see if I can remember the proof:

First off, this proof uses infinite series. Put basically, any differentiable function can be represented by the sum of an infinite series. No, I don't expect you to know what that means. Just accept that:

cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + . . .

sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + . . .

e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + . . .

It follows that:

e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

i*sin(x) = (i*x) - (i*x^3)/(3!) + (i*x^5)/(5!) - (i*x^7)/(7!) + . . .

Now something interesting happens when you add cos(x) and i*sin(x) together. Lemme see if I can illustrate this clearly:

.....cos(x) = 1..........- (x^2)/(2!) ......................+ (x^4)/(4!) .................- (x^6)/(6!) + . . .
+ i*sin(x) = ..(i*x)..................... - (i*x^3)/(3!) .................+ (i*x^5)/(5!) - . . .
---------------------------------------------------------------------------------------------------------------
....e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

Don't be discouraged if the idea of infinite series is unfamiliar to you - just take the series I gave for sin(x), cos(x), and e^x for granted, and everything else I show above follows from simple arithmetic on the series.

Anyway, this creates the general equation:

e^(i*x) = cos(x) + i*sin(x)

if x = PI, then:

e^(i*PI) = cos(PI) + i*sin(PI) = -1 + i*0 = -1

Oh, and shame on your math teacher for not knowing this proof :p

[Cenright]

(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.)
I know I am looking at this from outside the equation, but still...

[GuidoHunter]

Uh, what? Where does zero come into this and when is it illegal to multiply by zero?

--Guido

http://andy.mikee385.com

[Kefit]

Well, some people use zero to set up this identity as:

e^(i*PI) + 1 = 0

And then they sometimes use that to try and prove the existence of God due to the existence of such a streamlined relation between what are more or less the five most important constants in all of math.

Of course, that is not what Cenright was talking about; rather this is the real way that zero gets involved with this equation.

[Anticrombie0909]

*head explodes*

[Omeganitros]

Actually, I think I got a grasp on that.

However, I can see that me and Infinity will be having a frustrating relationship in the future.

[sleeplessdragn]

I heard about the mathematical proof of the existence of God, but i had NO clue that this equation was the root of that proof. That simply astounds me. I actually did a one eye brow raise when i read that.

[Kilgamayan]

Jesus, I couldn't have produced that and I'm a math major.

I'll have to remember that.

[Tasselfoot]

All I remember about that proof is that my fat ass math god buddy used to geek around talking about e^(i*Pi) instead of -1 any time -1 came up, or some such multiplier of that.

Needless to say, yes... he's a math god. and yes... he's a virgin. and yes... he weighs 300lbs.

And, with that, I go back to hating math math and back to loving finance math.

[blahblah18]

OK ITS REAL SIMPLE...
kefit has it right...
they're called taylor series... anyone? anyone? Bueller? calculus?

[GuidoHunter]

Pfft, they're nothing but MacLaruin serieseses centered about zero.

--Guido

http://andy.mikee385.com

[igotrhythm]

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

[banditcom]

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

lol... That's for stupid people who fail to realize that the -1 is still in there to counter the 1.

Anyways, nice proof Kefit. I totally forgot about the different series, even though they're really cool. Currently, I'm so sick of proofs. Most of them I feel like "this doesn't seem to prove anything" but they do.

[Braydz]

I actually understood all of that...

Is all of this uni/college level maths? I'm still in high school... Also, curious... Do those equations of sine and cosine work with both radians and degrees, or only one? Never seen sine and cosine shown that way before.

[GuidoHunter]

All calculus is done in radians, and yes, this is Cal II stuff.

--Guido

http://andy.mikee385.com

[hatakikakashi]

(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.)
I know I am looking at this from outside the equation, but still...

I always thought that the rule was that you can't divide by zero. 50 multpied by zero equals (zero) 50 divided by zero equals (your not allowed to do that)

I never understood why you can't do that. Also all this math is beyond me so I'll shut up now.

[GuidoHunter]

I always thought that the rule was that you can't divide by zero. 50 multpied by zero equals (zero) 50 divided by zero equals (your not allowed to do that)

I never understood why you can't do that. Also all this math is beyond me so I'll shut up now.

See my post after that.

If you want to understand why you can't divide by zero, just look at limits. Since you can always make a number smaller and smaller, you'll never end, which is why there is no answer to dividing by zero.

--Guido

http://andy.mikee385.com

[Braydz]

When I think of dividing by zero, I just think of the unit circle. tan(pi/2) is undefined as tan(x) = sin(x)/cos(x), which would result in tan(pi/2) = 1/0. Now, as far as I was taught, the tan function is basically the length of the hypotenuse from the origin to a tangent running vertically from the point (1,0). If the hypotenuse is going up vertically from (0,0), they will never meet, or somehow meet at infinity. That is why a graph of tan(x) has asymptotets at pi/2 adding or subtracting any normal number of pi. The graph flies upwards, approaching infinite. And when you think of it, there is no number of 0's that can go into any number. Is infinite a real number?

Sorry if I didn't make any sense, I am trying to remember something my specialist maths teacher told me at the end of last year. The fact that I am also a "young" maths student doesn't help, along with the fact that I have trouble explaining myself. If anyone can please correct my foggy memory here, please do.

[HamHack]

wow.i didn't retain any of that.hmm.

[jimerax]

I actually understood all of that...

Is all of this uni/college level maths? I'm still in high school... Also, curious... Do those equations of sine and cosine work with both radians and degrees, or only one? Never seen sine and cosine shown that way before.

It's definitely high school level math.

[bigsleytheoaf]

http://www.amazon.com/exec/obidos/tg/detail/-/0070542341/qid=1113127595/sr=8-1/ref=pd_csp_1/002-8038317-2967266?v=glance&s=books&n=507846

PS - Math is a model. The statement e^(pi * i) = -1 has to do with an analytic extension of the exponential function onto the complex plane. There's no way to "prove" it just like there's no way to "prove" the axiom of choice (well, they're kind of different - you can prove existence and uniqueness of such an extension...) Kind of like, you know how 3! = 6? Well you can extend it analytically to the gamma function so that you can "have statements like" (3/2)! = sqrt(pi) and definitions of factorials with non-integral negative domain. But they're not "true," they're just "defined extensions."

PPS - Talking about math casually is like talking about sex casually. It's the past time of virgins. I guess maybe it just gets me a little irritated when people are like "oh the wonder of math" when they haven't even studied basic Real Analysis. It also gets me irritated when people talk about physics without understanding it and chemistry without understanding it and everything without understanding it, where "understanding it" means getting a Ph.D. in it at a first or second-tier university. There's alot to understand out there, so much so that there's little use in pretending to understand. Blah blah blah I'm tired you suck donk.

PPPS - This is in reference to Kefit's "proof," by the way...

[hatakikakashi]

My head hurts I don't get it. I need more schooling.

[alainbryden]

where "understanding it" means getting a Ph.D. in it at a first or second-tier university.

Don't be an ass. Only uptight pricks think the only way to know what you're talking about is to have a standardised education.

What Kefit brilliantly showed is commonly known, in mathimatical terms, as a proof, so don't be putting quotations like he did something wrong.


Kefit, that was awesome. I envy your grasp of mathematics. I'm still working really hard to keep up with second term calc.

[bigsleytheoaf]

Why don't you "prove" that 1+1 = 2?
Can you do this?
A very careful inspection of what "proof" means will show you that Kufit didn't "prove" anything. Alot of people get hung up on silly things like:

0/0 = 1

???

You can define it that way, and you can "prove" it, but to do so requires that you change the whole of your mathematical apparatus around it.

Whatever.

Hey, isn't the theory of relativity awesome?
How about those quantum physics?

-BigsleyTheOaf

[Braydz]

Ok then, if you are such a know-all, why not share some of your knowledge with us?

[Kefit]

Hey, isn't the theory of relativity awesome?
How about those quantum physics?

-BigsleyTheOaf

Hi.

I can go on about both of those subjects, and have done it before on these boards.

Anyway, what is your point? Of course I didn't really "prove" anything - my proof only works in the modern and widely accepted model for both mathematics and numbers, both of which, from an objective standpoint, are fairly arbitrary. It is true that this proof doesn't work in other mathematical models, but I don't think that anyone here is really concerned with those.

Besides, it is impossible to prove anything absolutely in this world. It is generally accepted that when people say they are proving something, that they are proving it within a given widely accepted model. Will this cut it in Ph.D level math and physics? Probably not, but I don't need to worry about that yet.

[blahblah18]

hey Dicktard... I've proved that 1+1=2 .. it is a rather complicated proof actually, a little over 2 pages long. So yeah, try and grow a brain champ.

Also 0/0 does not = 1
0/0 = indeterminant, which means that it represents some answer, but that it can not be determined without other information, whereas
x/0 = undefined

Btw, I assume with safety that
A) you don't have a PHD at any university let alone a good one
B) you probably aren't in college
C) you're so much dumber then I am that I should just pity you.

[hatakikakashi]

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAH

[Loverofstories]

*SYSTEM OVERLOAD!!!!!!!!*

[hatakikakashi]

I got all worried when you guys made me think that 1+1 might not equal 2... So I did some research.


I took 1+1 apples then I preceded to count them. There were 2. Then following the scientific method I sent my findings to several different "researchers" They all were able to duplicate my test results, except for my cat Proffesor Kitty who after exhausting study came to the conclusion that there were meow apples.

[sleeplessdragn]

Guys, you have turned a perfectly legit question thread in the CT forum into stupid. BigsleyTheOaf being the source of this stupid by unnecessary flaming.

And my input on BigsleyTheOaf's first post:
It is impossible to measure the amount of understanding needed in order to explain anything. In fact I can just as well casually mention that I don't think you understand enough about math to explain it, because only people who have researched mathematics intensely at the Ph.D level for at least thirty years is even remotely educated enough. On the other hand, it can be said that I think if you have a basic grasp on the concpet of Algebra that you are educated enough to explain it. It is all relative.

Let this thread die now, it has served its purpose.

[blahblah18]

save for the need for more people to call this guy dicktard from now on... and seriously, the "concept" of addition is not one to be just taken for granted. Peano's Axioms are a good start if you're actually interested

[GuidoHunter]

I'm interested, but I could google it if you don't want to post it all.

--Guido

http://andy.mikee385.com

[bigsleytheoaf]

Hmm, I've written 2 responses, but they don't seem to want to post. Here's a summary:

1) I go to MIT, and I major in Math. I'm not a Ph.D, but I'm good at it. However, there are many people much better than me and my original post was not intended to be "bragging" although I can see how you could interpret it that way.

2) I was trying to express irritation at people who portray a "casual" knowledge of math because I think that this is a very irritating form of boasting. "Oh man, you know what's so awesome about the Radon-Nikodym Theorem!?!?"

Whatever, it doesn't matter.
You guys jump to alot of conclusions and sure start with the feisty comments really quick-like.

[FlashStinger]

what street is MIT on?

[bigsleytheoaf]

"MIT" is large, but the usual address that people associate with it is 77 Mass. Ave. I live at 70 Amherst Street, Cambridge.

[FlashStinger]

ok, can you disprove that this doesn't work

let a=b=1

a=a
a*a=ab
a^2=ab
-ab -ab
a^2 - ab = 0
a(a-b) = 0
/(a-b) /(a-b)
a = 0

Therefore 1= 0, according to the first statement a = b = 1

:)


how about more complex,
Let m = n+1
n+1 = n+1
(n+1)+(n+1) = m(n+1)
(n+1)^2 - m(n+1) = 0
(n+1)(n+1-m) = 0
/(n+1-m) /(n+1-m)

n+1 = 0

???? no matter what n equals, n+1 is always 0, therefore all numbers are pointless, and that degree you will be getting is semi-pointless. Not really. but i like to ask smart mathe people these questions. Let's see if you can fix it.

[FlashStinger]

oh, and for such a bright man, I don't recommend giving out my address on a forum. Hackers and such getting your info.... not normally good. If you didn't play ffr, I'd take you for a spin ;)

[talisman]

uh.... (n+1) + (n+1) doesn't equal m(n+1), if m is (n+1). That's like saying (n+1)(n+1) = (n+1) + (n+1).

and you can't divide by zero (a-b) in the first one.

[bigsleytheoaf]

re flash stinger - good luck with "getting my info." Does anyone want my IP? What would one do, exactly? All of my important documents are encrypted and backed up on athena.

Word on the street is that hacking is dead.

BTW: MIT's root pwd is public -> You can ssh remotely and use neo to find all a person's information if you have any other bit of their information... Hacking people at MIT isn't hard, it's just not done because it's poor form/immature.

---

Actually, someone used the root login to install a fake login screen on public workstations recently and harvested a bunch of email addresses. Too bad they got caught and their asses are getting expelled. *shrug*

-Bigsley

[FlashStinger]

uh.... (n+1) + (n+1) doesn't equal m(n+1), if m is (n+1). That's like saying (n+1)(n+1) = (n+1) + (n+1).

and you can't divide by zero (a-b) in the first one.
sorry mis typed....

(n+1)(n+1) = m(n+1)

and yes, that is the response, i divided by 0 in both. Dividing by zero leads to some pretty interesting results, eh? This shows you why it's important to observe rules about domain restrictions, or you can come up with all kinds of mayhem. (i.e. [a-b] and [n+1-m] both equal 0). I'm glad there are some smart people out there. Some people on other forums believed me and had to ask for why it works. re flash stinger - good luck with "getting my info." Does anyone want my IP? What would one do, exactly? All of my important documents are encrypted and backed up on athena.

Word on the street is that hacking is dead.

BTW: MIT's root pwd is public -> You can ssh remotely and use neo to find all a person's information if you have any other bit of their information... Hacking people at MIT isn't hard, it's just not done because it's poor form/immature.

---

Actually, someone used the root login to install a fake login screen on public workstations recently and harvested a bunch of email addresses. Too bad they got caught and their asses are getting expelled. *shrug*

-Bigsley

Sure most hacking is "dead", but it is still harmful. Just a few weeks ago a few hundred identities were stolen from a college up in California where they got the SS#'s of many people and credit card numbers. With that kinda of stuff, you can cause damage.

FYI, "Hacking" isn't dead. Cracking is on the drop, but most "hacks" are not reported, and are hardly ever harmful. Learn the lingo. Unfortunately, there will always be script kiddies attacking server's and games, just for fun. Hey, I have done it, but thinking that hacking is "dead" is simply stupid.

[bigsleytheoaf]

eh.

I may be stupid, but I'm still not really very worried. To the best of my knowledge, the following are true:

1) The russian mafia probably has all of my identifying information (ssn, credit card #, etc.)

2) I don't really have anything for anyone to steal. I don't have much money and if someone wanted it I would probably give it to them (assuming they had at least a coherent reason). If they want my few hundred dollars to buy a watch or something then that's just dumb.

-Bigsley

[FlashStinger]

true, Hacking is mostly done to small online buisnesses, who eventually fix the problem. So crackers inevitably help the world!!!! not really, but w/e

[blahblah18]

FlashStinger should be banned from CT

MIT was too depressing and brick and scary, so I didn't go there.. I also loved how sterotypical it was that a white man and an asian woman were having a jewish wedding outside when I went to visit it back in the day

[FlashStinger]

I should be banned from CT? on what grounds......
I am not claiming to be a hacker or cracker of any sorts and sorry if i stepped on your toes about some things, but as I recall this is a forum based on thinking critically and I have thought about my comments. I also in no way claim to be in any way smarter than a man or woman who can get into MIT. Heck, I'm just a highschool student in a school possibly losing it's acreditation so don't take many of my responses on Calc III that seriously. Sure I understand e^(i*π) but that doesn't mean much.

[banditcom]

I think what bigsleytheoaf was referring to are "wannabes". However, I don't think that was appropiate for this thread bigsley. I think everyone in here was just excited about what they know and felt good about it, especially since many who did are quite young. I can see your point, but it wasn't needed for this thread.

[Boochypa]

Since I'm insane enough to return to the original purpose of this thread, I'll offer an alternative "proof." (Note that y' = dy/dx).

y = cos(x) + isin(x)
y' = -sin(x) + icos(x)
y' = i(isin(x) + cos(x))
y' = iy
y'/y = i
ln(y) = ix
y = e^(ix)

e^(ix) = cos(x) + isin(x)

Plug in pi for x and you get your little -1 bit.

*Hopes that anyone who knows better will conveniently ignore my exclusion of an integration constant.*

[VxDx]

ok, I'd just like to make the point that bigsley (1) posted all about his MIT math major in one of his first 9 posts (2) is a member of FFR (3) is a MATH MAJOR (4) brags about this on the internet.

I'm willing to bet that at least one of the following is true.
- his rich parents pay for his amazing education
- he lacks social skills
- he's arrogant in real life too, though too timid to show it outwardly (which, admittedly, is probably a good thing)
- is arguing for the sake of being different